From the table, the value V = 5200 cm³ is between x = 21 cm and x = 22 cm.
Computing the average of the volumes associated to these x-values, we get:
V = (4,849.1 + 5,575.3)/2
V = 5212.2
which is near V = 5200 cm³. Then, the x-value related to V = 5200 cm³ is approximately the average between x = 21 and x = 22, that is:
x = (21 + 22)/2
x = 21.5 cm
